Is there a short way to implement Default for a struct that has a field that does not implement Default?

I have a struct that has 20 fields:

struct StructA {
    value1: i32,
    value2: i32,
    // ...
    value19: i32,
    day: chrono::NaiveDate,
}

I'd like to impl Default trait for StructA. I tried to add #[derive(Default)] to the struct, but chrono::NaiveDate doesn't implement Default.

I then tried to implement Default for StructA:

impl Default for StructA {
    fn default() -> Self {
        Self {
            value1: Default::default(),
            value2: Default::default(),
            // ...
            value19: Default::default(),
            day: chrono::NaiveDate::from_ymd(2021, 1, 1),
        }
    }
}

This code works fine, but the parts of value1 through value19 are redundant. Is there a solution with less code?

• I defined StructA to deserialize JSON data via serde-json so I can't change the struct's definition.
• A value of day: chrono::NaiveDate is always given from JSON data, so I want to avoid day: Option<chrono::NaiveDate>.

The derivative crate makes this kind of thing a breeze:

#[derive(Derivative)]
#[derivative(Default)]
struct StructA {
    value1: i32,
    value2: i32,
    // ...
    value19: i32,
    #[derivative(Default(value = "NaiveDate::from_ymd(2021, 1, 1)"))]
    day: NaiveDate,
}

If you want to avoid external crates, your options are:

• the approach you already used, with the downside that you must name all the fields. Also note that you don't need to repeat Default::default() for each numeric field, a simple 0 will work as well.
• make day an Option and derive Default, with the downside that it will default to None, bear a size penalty, and you'll have to use unwrap_or(0) at run-time to access it.
• make day a newtype that wraps NaiveDate and implements Default to set it to the desired value, with the downside that you'll need to access the underlying NaiveDate through a (zero-cost) field or method.